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Find the value of the following definite integral:

$$\int_0^{1}\frac{\arctan(t)}{1+t}dt$$

I tried using integration by parts, but it gives another complicated integral, so possibly the antiderivative does not exist. What other methods can I apply?

Shubham
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    Are you sure that the denominator is not $1+t^2$ instead of $1+t$ ? – Claude Leibovici Oct 07 '14 at 09:40
  • An antiderivative exists according to wolframalpha, but it's weird. Better try with definite integration methods. Where did you find it? Are you sure it has a closed form? – UserX Oct 07 '14 at 09:41
  • @Claude, yes of course. Otherwise the question is just a simple substitution – Shubham Oct 07 '14 at 09:43
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    $\arctan\dfrac\pi4$ doesn't make much sense. Perhaps you meant to write $\displaystyle\int_0^1$, since $\arctan1=\dfrac\pi4$ ? In which case, the answer is $\dfrac\pi8\ln2$. – Lucian Oct 07 '14 at 09:56
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    This integral is so difficult that I just wondered. – Claude Leibovici Oct 07 '14 at 09:58
  • I'm sorry, the upper limit is one. Will have to edit. – Shubham Oct 07 '14 at 09:59
  • Maple outputs $$ {\frac {7,i}{96}}{\pi }^{2}-{\it Catalan}/2+i{\it dilog} \left( 1/2, \sqrt {2}-i/2\sqrt {2}+1 \right) -1/8,\pi ,\ln \left( 2+\sqrt {2} \right) +i/4\pi ,\arctan \left( \left( 1+\sqrt {2} \right) ^{-1} \right) +1/4,\pi ,\ln \left( 2 \right) +i{\it dilog} \left( -1/2, \sqrt {2}+i/2\sqrt {2}+1 \right) -1/8,\pi ,\ln \left( 2-\sqrt {2} \right) -i/4\pi ,\arctan \left( \left( \sqrt {2}-1 \right) ^{-1} \right). $$ Its numerical value is $0.2721982609 $. – user64494 Oct 07 '14 at 10:04
  • @Lucian , yes the answer is indeed $\dfrac {\pi}{8} \ln 2$ our teacher told us so. I don't see how that comes out. – Shubham Oct 07 '14 at 10:08

2 Answers2

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Let us write $$I=\int_0^1\dfrac{\arctan t}{1+t}dt$$ The change of variables $t=\dfrac{1-x}{1+x}$ shows that $$\eqalign{I&=\int_0^1\arctan\left(\frac{1-x}{1+x}\right)\frac{dx}{1+x}\\ &=\int_0^1\left(\frac{\pi}{4}-\arctan x\right)\frac{dx}{1+x}\\ &=\frac{\pi}{4}\int_0^1\frac{dx}{1+x}-\int_0^1\frac{\arctan x}{1+x}dx\\ &=\frac{\pi}{4}\ln 2-I }$$ Thus $I=\dfrac{\pi}{8}\ln 2$.$\qquad \square$

Omran Kouba
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It is not as easy an integral as it first appears to be. I am cutting/pasting solutions in four cases obtained by Mathematica Version 8. ...including the upper limit 1 you changed.

First case result is real, even if it involves $i$ but fails to annul $i$ in the final expression.

Second case is what you asked for, evaluated numerically.

Third one actually involves Arctan, Log, and Catalan constant.

Fourth case with upper limit 1 has an easily obtainable result, but connection to simplification due the upper limit 1 is surprising...

$ \text{Integrate}[\text{ArcTan}[(u)]/(1+u),u] $

$ 1/32 (-5 I[\text{Pi}]{}^{\wedge}2+8 I[\text{Pi}] \text{ArcTan}[u]-32 I \text{ArcTan}$

$[u]{}^{\wedge}2-24[\text{Pi}] \text{Log}[2]+16[\text{Pi}] \text{Log}$

$[1+E{}^{\wedge}(-2 I \text{ArcTan}[u])] $

$-32 \text{ArcTan}[u] \text{Log} [1+E{}^{\wedge}(-2 I \text{ArcTan}[u])]+ $

$ 8[\text{Pi}] \text{Log}[1-I E{}^{\wedge}(2 I \text{ArcTan}[u])] $

$ +32 \text{ArcTan}[u] \text{Log}[1-I E{}^{\wedge}(2 I \text{ArcTan}[u])]+ $

$8[\text{Pi}] \text{Log}[1+u{}^{\wedge}2]-8[\text{Pi}] \text{Log}$ $[\text{Sin}$

$[[\text{Pi}]/4+ \text{ArcTan}[u]]]-16 I \text{PolyLog}[2,-E{}^{\wedge}(-2 I $

$\text{ArcTan}[u])]-16 I \text{PolyLog}[2,I E{}^{\wedge}(2 I \text{ArcTan}[u])]) $

$ \text{NIntegrate}[\text{ArcTan}[(u)]/(1+u),\{u,0,\text{Pi}/4.\}] $

0.189728

$ \text{Integrate}[\text{ArcTan}[(u)]/(1+u),\{u,0,\text{Pi}/4\}] $

$ 1/96 (-48 \text{Catalan}-47 I[\text{Pi}]{}^{\wedge}2+[\text{Pi}] (24 I$

$ \text{ArcTan}[[\text{Pi}]/4]-2$

$\text{Log}[4096]+24 (\text{Log}[1-I E{}^{\wedge}(2 I \text{ArcTan}$

$[[\text{Pi}]/4])]+2 \text{Log}[I/(-4 I+[\text{Pi}])]+\text{Log}[16+[\text{Pi}]$

${}^{\wedge}2]+\text{Log}[ \text{Csc}[[\text{Pi}]/4+ $

$ \text{ArcTan} [[\text{Pi}]/4]]]))-48 I (2 \text{ArcTan}[[\text{Pi}]/4] $

$ {}^{\wedge}2+2I \text{ArcTan}[[\text{Pi}]/4] (\text{Log}[-((1+I)/(4 $

$ I+[\text{Pi}]))]-\text{Log}[(8 I)/((-4 I+[\text{Pi}]) $

$ (4+[\text{Pi}]))])+\text{PolyLog}[2,-((4+I[\text{Pi}])/(4 $

$ I+[\text{Pi}]))]+\text{PolyLog}[2,(4 I+[\text{Pi}])/(-4 I+[\text{Pi}])])) $

$ \text{Integrate}[\text{ArcTan}[(u)]/(1+u),\{u,0,1\}] $

$ log (2) \pi/8 $

The given function is regular, is a product of two functions one increasing, other decreasing, so has a maximum, ( at t = 1.22913).

Narasimham
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  • In Mathematica, you can choose to copy output as Latex code. This makes it much more human-readable. – user111187 Oct 07 '14 at 15:14
  • Mathematica forever? Not quite in this case :p – yo' Oct 07 '14 at 16:10
  • Sorry, messed up with it first time me doing.. At least serves the purpose that not so nice integration limits invite Arctan Log, PolyLog, Catalan and what have you :) – Narasimham Oct 07 '14 at 16:15